Question

two identical insulated charged copper spheres a and b have their centre separated by a distance of 50 cm a third sphere of sme size but uncharged is brought in contact with the first then brought in contact with the second and finally removed from both what is the new force of repulsion between a and b? the radii of a and b are negligible compared to the distance of separation​

Answer 1

Answer:

QA=6.5×10−7

QB=6.5×10−7

Let the uncharged sphere be C.

QC=0

After contacting with A, the charge gets divided equally between A and C.

So, the new charges on A and C are 6.5×10−7/2 and 6.5×10−7/2, respectively.

Now when we brought into contact C and B, the total charge on B and C again gets divided equally between B and C.

So, the new charges on B and C are 23.25×10−7+6.5×10−7 and 23.25×10−7+6.5×10−7, respectively.

So now,

QA′=3.25×10−7

QB′=4.875×10−7

Electrostatic force now is F=QA′QB′/4πϵor2

=5.7×10−8 N